Trigonometry Examples

(- 2, 0)

$(- 2, 0)$ ,

(1, 7)

$(1, 7)$

Step 1

Use the dot product formula to find the angle between two vectors.

θ = arccos (\frac{a⃗ \cdot b⃗}{| a⃗ | | b⃗ |})

$θ = \arccos (\frac{a⃗ \cdot b⃗}{| a⃗ | | b⃗ |})$

Step 2

Find the dot product.

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Step 2.1

The dot product of two vectors is the sum of the products of the their components.

a⃗ \cdot b⃗ = - 2 \cdot 1 + 0 \cdot 7

$a⃗ \cdot b⃗ = - 2 \cdot 1 + 0 \cdot 7$

Step 2.2

Simplify.

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Step 2.2.1

Simplify each term.

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Step 2.2.1.1

Multiply

- 2

$- 2$ by

1

$1$ .

a⃗ \cdot b⃗ = - 2 + 0 \cdot 7

$a⃗ \cdot b⃗ = - 2 + 0 \cdot 7$

Step 2.2.1.2

Multiply

0

$0$ by

7

$7$ .

a⃗ \cdot b⃗ = - 2 + 0

$a⃗ \cdot b⃗ = - 2 + 0$

a⃗ \cdot b⃗ = - 2 + 0

$a⃗ \cdot b⃗ = - 2 + 0$

Step 2.2.2

Add

- 2

$- 2$ and

0

$0$ .

a⃗ \cdot b⃗ = - 2

$a⃗ \cdot b⃗ = - 2$

a⃗ \cdot b⃗ = - 2

$a⃗ \cdot b⃗ = - 2$

a⃗ \cdot b⃗ = - 2

$a⃗ \cdot b⃗ = - 2$

Step 3

Find the magnitude of

a⃗

$a⃗$ .

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Step 3.1

The norm is the square root of the sum of squares of each element in the vector.

| a⃗ | = \sqrt{{(- 2)}^{2} + 0^{2}}

$| a⃗ | = \sqrt{{(- 2)}^{2} + 0^{2}}$

Step 3.2

Simplify.

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Step 3.2.1

Raise

- 2

$- 2$ to the power of

2

$2$ .

| a⃗ | = \sqrt{4 + 0^{2}}

$| a⃗ | = \sqrt{4 + 0^{2}}$

Step 3.2.2

Raising

0

$0$ to any positive power yields

0

$0$ .

| a⃗ | = \sqrt{4 + 0}

$| a⃗ | = \sqrt{4 + 0}$

Step 3.2.3

Add

4

$4$ and

0

$0$ .

| a⃗ | = \sqrt{4}

$| a⃗ | = \sqrt{4}$

Step 3.2.4

Rewrite

4

$4$ as

2^{2}

$2^{2}$ .

| a⃗ | = \sqrt{2^{2}}

$| a⃗ | = \sqrt{2^{2}}$

Step 3.2.5

Pull terms out from under the radical, assuming positive real numbers.

| a⃗ | = 2

$| a⃗ | = 2$

| a⃗ | = 2

$| a⃗ | = 2$

| a⃗ | = 2

$| a⃗ | = 2$

Step 4

Find the magnitude of

b⃗

$b⃗$ .

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Step 4.1

The norm is the square root of the sum of squares of each element in the vector.

| b⃗ | = \sqrt{1^{2} + 7^{2}}

$| b⃗ | = \sqrt{1^{2} + 7^{2}}$

Step 4.2

Simplify.

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Step 4.2.1

One to any power is one.

| b⃗ | = \sqrt{1 + 7^{2}}

$| b⃗ | = \sqrt{1 + 7^{2}}$

Step 4.2.2

Raise

7

$7$ to the power of

2

$2$ .

| b⃗ | = \sqrt{1 + 49}

$| b⃗ | = \sqrt{1 + 49}$

Step 4.2.3

Add

1

$1$ and

49

$49$ .

| b⃗ | = \sqrt{50}

$| b⃗ | = \sqrt{50}$

Step 4.2.4

Rewrite

50

$50$ as

5^{2} \cdot 2

$5^{2} \cdot 2$ .

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Step 4.2.4.1

Factor

25

$25$ out of

50

$50$ .

| b⃗ | = \sqrt{25 (2)}

$| b⃗ | = \sqrt{25 (2)}$

Step 4.2.4.2

Rewrite

25

$25$ as

5^{2}

$5^{2}$ .

| b⃗ | = \sqrt{5^{2} \cdot 2}

$| b⃗ | = \sqrt{5^{2} \cdot 2}$

| b⃗ | = \sqrt{5^{2} \cdot 2}

$| b⃗ | = \sqrt{5^{2} \cdot 2}$

Step 4.2.5

Pull terms out from under the radical.

| b⃗ | = 5 \sqrt{2}

$| b⃗ | = 5 \sqrt{2}$

| b⃗ | = 5 \sqrt{2}

$| b⃗ | = 5 \sqrt{2}$

| b⃗ | = 5 \sqrt{2}

$| b⃗ | = 5 \sqrt{2}$

Step 5

Substitute the values into the formula.

θ = arccos ⎛ ⎜ ⎝ \frac{- 2}{2 (5 \sqrt{2})} ⎞ ⎟ ⎠

$θ = \arccos (\frac{- 2}{2 (5 \sqrt{2})})$

Step 6

Simplify.

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Step 6.1

Cancel the common factor of

- 2

$- 2$ and

2

$2$ .

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Step 6.1.1

Factor

2

$2$ out of

- 2

$- 2$ .

θ = arccos ⎛ ⎜ ⎝ \frac{2 \cdot - 1}{2 (5 \sqrt{2})} ⎞ ⎟ ⎠

$θ = \arccos (\frac{2 \cdot - 1}{2 (5 \sqrt{2})})$

Step 6.1.2

Cancel the common factors.

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Step 6.1.2.1

Cancel the common factor.

$θ = \arccos (\frac{2 \cdot - 1}{2 (5 \sqrt{2})})$

Step 6.1.2.2

Rewrite the expression.

$θ = \arccos (\frac{- 1}{5 \sqrt{2}})$

Step 6.2

Move the negative in front of the fraction.

$θ = \arccos (- \frac{1}{5 \sqrt{2}})$

Step 6.3

Multiply

$\frac{1}{5 \sqrt{2}}$ by

$\frac{\sqrt{2}}{\sqrt{2}}$ .

$θ = \arccos (- (\frac{1}{5 \sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}}))$

Step 6.4

Combine and simplify the denominator.

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Step 6.4.1

Multiply

$\frac{1}{5 \sqrt{2}}$ by

$\frac{\sqrt{2}}{\sqrt{2}}$ .

$θ = \arccos (- \frac{\sqrt{2}}{5 \sqrt{2} \sqrt{2}})$

Step 6.4.2

Move

$\sqrt{2}$ .

$θ = \arccos (- \frac{\sqrt{2}}{5 (\sqrt{2} \sqrt{2})})$

Step 6.4.3

Raise

$\sqrt{2}$ to the power of

$1$ .

$θ = \arccos (- \frac{\sqrt{2}}{5 ({\sqrt{2}}^{1} \sqrt{2})})$

Step 6.4.4

Raise

$\sqrt{2}$ to the power of

$1$ .

$θ = \arccos (- \frac{\sqrt{2}}{5 ({\sqrt{2}}^{1} {\sqrt{2}}^{1})})$

Step 6.4.5

Use the power rule

$a^{m} a^{n} = a^{m + n}$ to combine exponents.

$θ = \arccos (- \frac{\sqrt{2}}{5 {\sqrt{2}}^{1 + 1}})$

Step 6.4.6

Add

$1$ and

$1$ .

$θ = \arccos (- \frac{\sqrt{2}}{5 {\sqrt{2}}^{2}})$

Step 6.4.7

Rewrite

${\sqrt{2}}^{2}$ as

$2$ .

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Step 6.4.7.1

Use

$\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite

$\sqrt{2}$ as

$2^{\frac{1}{2}}$ .

$θ = \arccos (- \frac{\sqrt{2}}{5 {(2^{\frac{1}{2}})}^{2}})$

Step 6.4.7.2

Apply the power rule and multiply exponents,

${(a^{m})}^{n} = a^{m n}$ .

$θ = \arccos (- \frac{\sqrt{2}}{5 \cdot 2^{\frac{1}{2} \cdot 2}})$

Step 6.4.7.3

Combine

$\frac{1}{2}$ and

$2$ .

$θ = \arccos (- \frac{\sqrt{2}}{5 \cdot 2^{\frac{2}{2}}})$

Step 6.4.7.4

Cancel the common factor of

$2$ .

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Step 6.4.7.4.1

Cancel the common factor.

$θ = \arccos (- \frac{\sqrt{2}}{5 \cdot 2^{\frac{2}{2}}})$

Step 6.4.7.4.2

Rewrite the expression.

$θ = \arccos (- \frac{\sqrt{2}}{5 \cdot 2^{1}})$

Step 6.4.7.5

Evaluate the exponent.

$θ = \arccos (- \frac{\sqrt{2}}{5 \cdot 2})$

Step 6.5

Multiply

$5$ by

$2$ .

$θ = \arccos (- \frac{\sqrt{2}}{10})$

Step 6.6

Evaluate

$\arccos (- \frac{\sqrt{2}}{10})$ .

$θ = 98.13010235$